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An excellent puzzle

July 18, 2017 3 comments

Today I saw this tweet:

The puzzle looked grand. Thanks to those people that tweeted at me to make sure if seen it, it’s much appreciated.

The puzzle itself is:

I drew it out and labelled a few things:

But soon realised that it’s impossible unless you make assumptions. 

With the assumption that the vertex of the triangle is at the midpoint if the line I was in a position to have a good crack at it. My first thought, as is often the case, was to run at it using right angled triangles:

My initial thought was to use right angled trigonometry, but I realised I’d probably need to approximate or use some maths software and that would take a bit of the fun out of it. I presumed I’d be able to find an exact answer in a better way.

I realised the big triangle and the green triangle were similar and I could easily work out the area of the big triangle.

I then realised I didn’t have the scale factor. I went back to rats.

Then realised I had another similar isoceles triangle to play with:

Using similarity I found the “base” of DFG and used that to find length EG. Thus giving me a scale factor between the blue and the green triangle.

As mentioned previously I knew that rt2/(SF^2) was the green area so using the scale factor of 3 I got the required area to be rt2/9. 

I would like to say that’s what I did. That’s what I see I should have done while writing this up. But it’s not what I did at the time. I took a longer way round. I got giddy with triangles:

Used  Pythagoras’s Theorem to find the peep height and found the area that way.

Luckily I got the same answer.

I then saw the same tweeter had tweeted this:

This is the same question but altered slightly in the information given and what is required as the final solution. If you make the same assumption it follows from the tan ratio that all the distances are the same, so you need to do the same to that point and then find the ratio green area / blue area. I’d done most of it above, so I finished it off:

I love this puzzle, and I hope to use it in my classes next year. I may give it to year 12 tomorrow and see if they can crack it. I think I prefer the second variation. I’d love to hear your thoughts on it, and how you solved it. Let me know in the comments or via social media. 

An interesting area puzzle 

July 5, 2017 11 comments

Regular readers will know that I have a tendency to collect puzzles and I like to have a go at them. This evening I had a crack at this while my daughter was playing before bed.

It looked interesting when I saw it on Facebook a while ago and so I thought I’d have a crack. If you haven’t yet, do it now. I want to know if you took the same approach!

I wasn’t too sure where to start, so I drew it out, labelled some stuff and came up with some equations:

I thought if I multiples two of my equations together and rearranged I could get the yellow area as the subject:

Then I needed dw:

Then I needed cx:


I thought “now I need ay”, then realised I had it:

This meant I could.sub back in through the equations:

So the area:

I thought 27 was a nice answer, and I’m fairly sure it’s correct, however I have a feeling that I may have missed something blindingly obvious that would have gotten me there much quicker. 

If you did it, I’d love to hear your approach, especially if you spotted something I missed!

A nice little triangle puzzle

December 5, 2016 8 comments

Here is a nice little puzzle I saw from brilliant.org on Facebook.

Have you worked it out yet?

Here’s what I did:

First I drew a diagram (obviously).

And worked out the area of the triangle.

Then the area of each sector.


Leaving a subtraction to finish.

Categories: #MTBoS, Maths, Starters Tags: , ,

Circles puzzle

October 13, 2016 Leave a comment

Here’s a lovely puzzle I saw on Brilliant.org this week:

It’s a nice little workout. I did it entirely in my head and that is my challenge to you. Do it, go on. Do it now….

Scroll down for my answer….

Have you done it? You better have…..
I looked at this picture and my frat thought was that the blue and gold areas are congruent. Thus the entire picture has an area of 70. There are 4 overlaps, each has an area of 5, so the total area of 5 circles is 90. Leaving each circle having an area of 18.

This is a nice mental work out and I feel it could build proprtional reasoning skills in my students. I am hoping to try it on some next week.

Did you manage the puzzle? Did you do it a different way?

This post was cross posted to better questions here.

Categories: Maths, Starters Tags: , , , ,

A lovely simple trigonometry puzzle

May 8, 2016 4 comments

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

image

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

Cross-posted to Betterqs here.

May 8, 2016 Leave a comment

Sometimes a puzzle can look complicated,  but be rather simple (see this geometry puzzle). I love puzzles like this and I particularly like to test them out on classes to try and build their problem solving ability.

Just now, I saw the following trig puzzle from brilliant.org and I love it! It’s amazing!

image

Have you done it yet?

How long did it take you to spot it?

My initial thought was, it’s got three terms,  it’s bound to be a disguised quadratic that will factorise. A few seconds later I realised that it wasn’t. I saw the – sin^4 and suspected a difference of two squares but then a few seconds later it became clear.

If you haven’t spotted it yet, have a look at the expression rearranged:

Sin^6 + sin^4 cos^2 – sin^4

See it now? What if I rewrite it as:

Sin^4 sin^2 + sin^4 cos^2 – sin^4

I’m sure you have seen it now, but to be complete,  take the common factor of the first two terms:

Sin^4 (sin^2 + cos^2) – sin^4

Obviously sin^2 + cos^2 = 1, so we’re left with:

Sin^4 – sin^4 = 0

A lovely, satisfying, simple answer to a little brain teaser. Hope you liked it as much as I did.

A lovely angle puzzle

March 24, 2016 4 comments

I’ve written before about the app “Brilliant“, which is well worth getting, and I also follow their Facebook page which provides me with a regular stream questions. Occasionally I have to think about how to tackle them, and they’re excellent. More often, a question comes up that I look at and think would be awesome to use in a lesson.

Earlier this week this question popped up:

image

What a lovely question that combines algebra and angle reasoning! I can’t wait to teach this next time, and I am planning on using this as a starter with my y11 class after the break.

The initial question looks simple, it appears you sum the angles and set it equal to 360 degrees, this is what I expect my class to do. If you do this you get:

7x + 2y + 6z – 20 = 360

7x + 2y + 6z = 380 (1)

I anticipate some will try to give up at this point, but hopefully the resilience I’ve been trying to build will kick in and they’ll see they need more equations. If any need a hint I will tell them to consider vertically opposite angles. They should then get:

2x – 20 = 2y + 2z (2)

And

3x = 2x + 4z (3)

I’m hoping they will now see that 3 equations and 3 unknowns is enough to solve. There are obviously a number of ways to go from here. I would rearrange equation 3 to get:

x = 4z (4)

Subbing into 2 we get:

8z – 20 = 2y + 2z

6z = 2y + 20 (5)

Subbing into 1

28z + 2y + 6z = 380

34z = 380 – 2y (6)

Add equation  (5) to (6)

40z = 400

z = 10 (7)

Then equation 4 gives:

x = 40

And equation 2 gives:

60 = 2y + 20

40 = 2y

y = 20.

From here you can find the solution x + y + z = 40 + 20 + 10 = 70.

A lovely puzzle that combines a few areas and needs some resilience and perseverance to complete. I enjoyed working through it and I’m looking forward to testing it out on some students.

Cross-posted to Betterqs here.